PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 139, Number 10, October 2011, Pages 3719–3726
S 0002-9939(2011)10822-0
Article electronically published on February 24, 2011
ON THE EQUALITY CONDITIONS
OF THE BRUNN-MINKOWSKI THEOREM
DANIEL A. KLAIN
(Communicated by Thomas Schlumprecht)
Abstract. This article describes a new proof of the equality condition for the
Brunn-Minkowski inequality. The Brunn-Minkowski Theorem asserts that, for
compact convex sets K, L ⊆ Rn , the n-th root of the Euclidean volume Vn is
concave with respect to Minkowski combinations; that is, for λ ∈ [0, 1],
Vn ((1 − λ)K + λL)1/n ≥ (1 − λ)Vn (K)1/n + λVn (L)1/n .
The equality condition asserts that if K and L both have positive volume, then
equality holds for some λ ∈ (0, 1) if and only if K and L are homothetic.
Denote n-dimensional Euclidean space by Rn . Given compact convex subsets
K, L ⊆ Rn and a, b ≥ 0, denote
aK + bL = {ax + by | x ∈ K and y ∈ L}.
An expression of this form is called a Minkowski combination or Minkowski sum.
Since K and L are convex sets, the set aK + bL is also convex. Convexity also
implies that aK + bK = (a + b)K for all a, b ≥ 0, although this does not hold for
general sets. Two sets K and L are homothetic if K = aL + x for some a > 0 and
some point x ∈ Rn . The n-dimensional (Euclidean) volume of K will be denoted
by Vn (K).
The Brunn-Minkowski Theorem asserts that the n-th root of the Euclidean volume Vn is concave with respect to Minkowski combinations; that is, for λ ∈ [0, 1],
(1)
Vn ((1 − λ)K + λL)1/n ≥ (1 − λ)Vn (K)1/n + λVn (L)1/n .
If K and L have non-empty interiors, then equality holds for some λ ∈ (0, 1) if and
only if K and L are homothetic. This article describes a new proof of this equality
condition, using a homothetic projection theorem of Hadwiger.
The Brunn-Minkowski Theorem is the centerpiece of modern convex geometry
[1, 6, 14, 15]. This theorem encodes as special cases the classical isoperimetric
inequality (relating volume and surface area [14, p. 318]), Urysohn’s inequality
(relating volume and mean width, and strengthening the isodiametric inequality
[14, p. 318]), and families of inequalities relating mean projections [14, p. 333].
The concavity implied by (1) leads to families of second-order discriminant-type
inequalities for mixed volumes, such as Minkowski’s second mixed volume inequality [14, p. 317] and (after substantial additional labor) the Alexandrov-Fenchel
Received by the editors May 9, 2010 and, in revised form, September 2, 2010.
2010 Mathematics Subject Classification. Primary 52A20, 52A38, 52A39, 52A40.
c
2011
American Mathematical Society
Reverts to public domain 28 years from publication
3719
3720
DANIEL A. KLAIN
inequality [14, p. 327], a difficult and far-reaching result with consequences in geometric analysis, combinatorics, and algebraic geometry [4]. Analytic generalizations
of the Brunn-Minkowski theorem include the Prékopa-Leindler inequality [3, 10].
The Brunn-Minkowski theorem also serves as the starting point for analogous developments such as the dual Brunn-Minkowski theory for star-shaped sets [11], the
Lp -Brunn-Minkowski theory [12], capacitary Brunn-Minkowski inequalities [2], and
Brunn-Minkowski inequalities for integer lattices [7]. Of special interest are the
equality conditions for (1), which imply, for example, the uniqueness of solutions
to the Minkowski problem relating convex bodies to measures on the unit sphere
[1, 14]. A recent and comprehensive survey on the Brunn-Minkowski inequality and
its variations, applications, extensions, and generalizations can be found in [5].
There are many ways to prove the Brunn-Minkowski Theorem. For the case of
compact convex sets, Kneser and Süss used an induction argument on dimension
via slicing [14, p. 310]. This proof first verifies the inequality (1) and then addresses
the equality case with a more subtle argument.
Hadwiger and Ohmann give a more general proof by using the concavity of the
geometric mean to prove (1) for the case of rectangular boxes and follow with
a divide-and-conquer argument that extends (1) to finite unions of boxes. They
conclude with an approximation step that verifies (1) for all measurable sets [9]
(see also [5] and [16, p. 297]). Hadwiger and Ohmann also show that if equality
holds in (1), then K and L must both be compact convex sets with at most a set
of measure zero removed. Therefore, the question of equality in (1) is addressed
completely (up to measure zero) by the case of compact convex sets.
Another especially intuitive proof of the inequality (1) for compact convex sets
uses Steiner symmetrization [5, 16]; however, this method relies on approximation
and gives no insight into the equality conditions.
In contrast to earlier methods, the proof of the equality condition for (1) presented in this note uses orthogonal projection rather than slicing. Sections 1, 2, and
3 provide some technical background. The new proof is presented in Section 4.
1. Background
Let Kn denote the set of compact convex subsets of Rn . If u is a unit vector
in Rn , denote by Ku the orthogonal projection of a set K onto the subspace u⊥ .
More generally, if ξ is a d-dimensional subspace of Rn , denote by Kξ the orthogonal
projection of a set K onto the subspace ξ. The boundary of a compact convex set
K relative to its affine hull will be denoted by ∂K.
Let hK : Rn → R denote the support function of a compact convex set K; that
is,
hK (v) = max x · v.
x∈K
The standard separation theorems of convex geometry imply that the support
function hK characterizes the body K; that is, hK = hL if and only if K = L.
If ξ is a subspace of Rn , then the support function of Kξ within the subspace
ξ is given by the restriction of hK to ξ. Support functions satisfy the identity
haK+bL = ahK + bhL . (See, for example, any of [1, 14, 16].)
If u is a unit vector in Rn , denote by K u the support set of K in the direction
of u; that is,
K u = {x ∈ K | x · u = hK (u)}.
EQUALITY CONDITIONS OF THE BRUNN-MINKOWSKI THEOREM
3721
If P is a convex polytope, then P u is the maximal face of P having u in its outer
normal cone.
Given K, L ∈ Kn and ε > 0, the function Vn (K +εL) is a polynomial in ε, whose
coefficients are given by Steiner’s formula [1, 14, 16]. In particular, the following
derivative is well defined:
d Vn (K + εL) − Vn (K)
=
(2)
Vn (K + εL).
nVn−1,1 (K, L) = lim
ε→0
ε
dε ε=0
The expression Vn−1,1 (K, L) is an example of a mixed volume of K and L. Since the
volume of any set is invariant under translation, it follows from the definition (2)
that, for any point p ∈ Rn ,
Vn−1,1 (K + p, L) = Vn−1,1 (K, L + p) = Vn−1,1 (K, L).
If P is a polytope, then the mixed volume Vn−1,1 (P, K) satisfies the classical
“base-height” formula
1 (3)
Vn−1,1 (P, K) =
hK (u)Vn−1 (P u ),
n
u⊥∂P
where this sum is finite, taken over all outer unit normals u to the facets on the
boundary ∂P . Since compact convex sets can be approximated (in the Hausdorff
metric) by convex polytopes, the equation (3) implies that, for K, L, M ∈ Kn and
a, b ≥ 0,
• Vn−1,1 (K, L) ≥ 0,
• Vn−1,1 (K, K) = Vn (K),
• Vn−1,1 (aK, bL) = an−1 bVn−1,1 (K, L),
• Vn−1,1 (K, L + M ) = Vn−1,1 (K, L) + Vn−1,1 (K, M ),
where the final identity follows from (3) and the linearity of support functions
with respect to Minkowski sums. For n ≥ 3, the function Vn−1,1 is not typically
symmetric in its parameters: usually Vn−1,1 (K, L) = Vn−1,1 (L, K). In particular,
the fourth (linearity) property above does not typically hold for the first parameter
of Vn−1,1 .
Important special (and more well-known) cases of mixed volumes result from
suitable choices of K or L. For example, if B is the unit ball, centered at the
origin, then (3) implies that nVn−1,1 (P, B) gives the surface area of P . A limiting
argument then yields the same fact for nVn−1,1 (K, B), where K is any compact
convex set. Similar arguments imply that, if ou denotes the line segment with
endpoints at o and a unit vector u, then
(4)
nVn−1,1 (K, ou) = Vn−1 (Ku ),
the (n − 1)-volume of the corresponding orthogonal projection of K. These and
many other properties of convex bodies and mixed volumes are described in each
of [1, 14, 16].
One especially intuitive proof of the Brunn-Minkowski inequality (1) for compact
convex sets uses Steiner symmetrization. Given a unit vector u, view K as a
family of line segments parallel to u. Slide these segments along u so that each is
symmetrically balanced around the hyperplane u⊥ . By Cavalieri’s principle, the
volume of K is unchanged. Call the new set stu (K). It is not difficult to show
that stu (K) is also convex, and that stu (K) + stu (L) ⊆ stu (K + L). A little more
work verifies the following intuitive assertion: if you iterate Steiner symmetrization
3722
DANIEL A. KLAIN
of K through a suitable sequence of directions, these iterations tend to round out
the body K to a Euclidean ball BK having the same volume as the original set
K. Meanwhile, it follows from the aforementioned superadditivity relation that
BK + BL ⊆ BK+L , so that
Vn (K + L)1/n = Vn (BK+L )1/n ≥ Vn (BK + BL )1/n
= Vn (BK )1/n + Vn (BL )1/n = Vn (K)1/n +Vn (L)1/n .
The technical details behind Steiner symmetrization and the proof outlined above
can be found in [16, pp. 306-314]. Once again, because of an approximation step
(taking the limit of a sequence of Steiner symmetrizations), it is not clear from this
proof exactly when equality would hold in (1).
This matter is addressed by the following theorem.
Theorem 1.1 (Minkowski). If K and L are compact convex sets with non-empty
interiors, then equality holds in (1) if and only if K and L are homothetic.
While homothety is evidently sufficient, its necessity is far from obvious.
Simple arguments show that the Brunn-Minkowski inequality (1) is equivalent
to Minkowski’s mixed volume inequality,
(5)
Vn−1,1 (K, L)n ≥ Vn (K)n−1 Vn (L),
where the equality conditions are the same as for (1).
Note that (5) is trivial if either Vn (K) = 0 or Vn (L) = 0. Moreover, both sides
of (5) are positively homogeneous of degree n(n − 1) with respect to scaling the
body K and positively homogeneous of degree n with respect to scaling the body
L. It follows that, for K and L with non-empty interiors, the inequality (5) is
equivalent to the assertion that
Vn−1,1 (K, L) ≥ 1
(6)
whenever Vn (K) = Vn (L) = 1.
To prove (6), and the equivalent (5), using (1), suppose that Vn (K) = Vn (L) = 1,
and let
t
f (t) = Vn ((1 − t)K + tL) = (1 − t)n Vn K + 1−t
(7)
L ,
for t ∈ [0, 1). By (2) and the chain rule,
(8)
f (0) = −nVn (K) + nVn−1,1 (K, L) = −n + nVn−1,1 (K, L).
Since f 1/n is concave by (1), we have f ≥ 1 on [0, 1), while f (0) = 1, so that
f (0) ≥ 0, and Vn−1,1 (K, L) ≥ 1.
If equality holds in (1) for any value of t ∈ (0, 1), the concavity of f 1/n implies
that f 1/n must be linear. Since f (0) = f (1) = 1 by construction, the function f 1/n
is constant. It follows that f (0) = 0, so that Vn−1,1 (K, L) = 1 and equality holds
in (6).
To prove (1) using (5) the argument is even simpler. Denote Kt = (1 − t)K + tL.
Then
Vn (Kt ) = Vn−1,1 (Kt , Kt ) = (1 − t)Vn−1,1 (Kt , K) + tVn−1,1 (Kt , L)
≥ (1 − t)Vn (Kt )
by two applications of (5).
n−1
Vn (Kt ) n .
n−1
n
1
V (K) n + tVn (Kt )
n−1
n
1
V (L) n ,
The inequality (1) then follows after division by
EQUALITY CONDITIONS OF THE BRUNN-MINKOWSKI THEOREM
3723
Suppose equality holds in (5), or equivalently, in (6). The non-negative function
f 1/n defined by (7) is concave on the interval [0, 1] by (1). Since f (0) = f (1) = 1,
the function f 1/n ≥ 1 on [0, 1] by concavity. But equality in (6), together with the
identity (8), implies that f 1/n has zero derivative at t = 0. By concavity again, it
follows that f 1/n has a non-positive derivative and must be non-increasing on [0, 1].
In other words, f 1/n is constant and equality holds in (1).
For a more complete discussion, see any of [1, 5, 14, 16].
2. Mixed area
Denote the special case of 2-dimensional volume V2 by A for area, and denote
V1,1 (K, L) by A(K, L), the mixed area. Unlike the higher dimensional mixed volumes, the mixed area is symmetric in its parameters: A(K, L) = A(L, K). If Δ is
a triangle with outward edge unit normals u1 , u2 , u3 , then
1
(9)
A(K, Δ) = A(Δ, K) =
hK (ui )|Δui |,
2 i
where |Δui | denotes the length of the i-th edge of the triangle. This identity leads
to the following proposition.
Proposition 2.1. Let K, L ∈ K2 , and suppose that
(10)
A(K, Δ) = A(L, Δ)
for every triangle Δ in R2 . Then K and L are translates.
Proof. Translate K and L so that both lie in the first quadrant of R2 and are
supported by the coordinate axes. If e1 and e2 respectively denote the unit vectors
along the two coordinate axes, we now have hK (−e1 ) = hK (−e2 ) = 0, and similarly
for L. Since mixed area is invariant under translation, the identity (10) still holds
for every triangle Δ.
If u is a unit vector with positive coordinates, let Δ be a right triangle having
outward unit normals −e1 , −e2 , u. Since hK (−ei ) = hL (−ei ) = 0, it follows from (9)
and (10) that hK (u) = hL (u).
If u is a unit vector in one of the other three quadrants, a similar argument is
then made (using a triangle with unit normals u, one of the −ei ’s, and a suitable
choice from the first quadrant) to show that hK (u) = hL (u) once again. It follows
that hK = hL , so that K = L after the initial translations of K and L into the
positive quadrant.
3. Bodies with homothetic projections are homothetic
In Section 4 we give a proof of the equality case for (1) and (5) using a projection
argument that relies on the following elementary theorem of Hadwiger [8].
Theorem 3.1 (The Homothetic Projection Theorem). Suppose that K, L ∈ Kn
have non-empty interiors, where n ≥ 3. If Ku and Lu are homothetic for all unit
vectors u, then K and L are homothetic as well.
For completeness of presentation, here is an elementary proof of Theorem 3.1
due to Rogers [13].
3724
DANIEL A. KLAIN
Proof. Let ei = (0, . . . , 1, . . . , 0) denote the unit vector having unit ith coordinate,
for i = 1, . . . , n.
Translate and scale K and L so that both sets are supported by the positive
+
coordinate halfspace (e⊥
n ) , and moreover so that Ken = Len . The latter is possible,
because we are given that Ken and Len are initially homothetic.
It remains to show that, after these translations and dilations, we have K = L.
To show this, let u ∈ e⊥
n be a unit vector. Recall from the hypothesis of the
theorem that Ku = aLu + v for some a > 0 and some v ∈ u⊥ , where both a and v
are dependent on u.
⊥
Let ξ = Span{u, en }⊥ = u⊥ ∩ e⊥
n . Since ξ ⊆ en and Ken = Len , it follows that
⊥
Lξ = Kξ . Since ξ ⊆ u as well, we also have Kξ = aLξ + vξ , so that Lξ = aLξ + vξ .
Since n ≥ 3, we have dim ξ = n − 2 ≥ 1. Let Vn−2 denote volume in Rn−2 . Since
translation does not change volume,
Vn−2 (Lξ ) = Vn−2 (aLξ + vξ ) = an−2 Vn−2 (Lξ ).
Since L has non-empty interior, Vn−2 (Lξ ) > 0. It follows that a = 1 and vξ = o.
This implies that v ∈ ξ ⊥ = Span{u, en }; that is, v = bu + cen for some b, c ∈ R.
Moreover v · u = 0, since we assumed v ∈ u⊥ to begin with. It follows that v = cen
for some c ∈ R.
+
The positive coordinate halfspace (e⊥
n ) supports both K and L, so that hK (−en )
= hL (−en ) = 0. Since Ku = Lu + v and −en ∈ u⊥ , we have
0 = hK (−en ) = hKu (−en ) = hLu (−en ) + v · (−en ) = hL (−en ) − c = −c,
so that v = cen = 0, and Ku = Lu .
We have shown that Ku = Lu for all u ∈ e⊥
n . If w is a unit vector, then
⊥
⊥
⊥
w ∩ e⊥
n = ∅, so w ∈ u for some u ∈ en . It now follows that
hK (w) = hKu (w) = hLu (w) = hL (w).
In other words, hK (w) = hL (w) for every unit vector w, so that K = L.
4. Conditions for equality
We now have the tools to verify the equality condition for (1) and (5).
Proof of the Equality Condition. For λ ∈ [0, 1], denote Kλ = (1 − λ)K + λL. Suppose that equality holds in (1) for some λ0 ∈ (0, 1), where K and L have non-empty
interiors. Since equality holds for (1) if and only if equality holds for (5) and (6),
the homogeneity of mixed volumes allows us to assume without loss of generality
that
Vn (K) = Vn (L) = Vn (Kλ0 ) = Vn−1,1 (Kλ0 , L) = 1.
1/n
The concavity of Vn
(11)
then implies that
Vn (Kλ ) = 1 for all λ ∈ [0, 1].
Fix a value of λ ∈ [0, 1], and suppose M ∈ Kn is such that Vn−1,1 (Kλ , M ) ≤ 1.
Since Vn−1,1 (·, ·) is Minkowski linear in its second parameter, we have
Vn−1,1 (Kλ , (1 − x)L + xM ) = (1 − x)Vn−1,1 (Kλ , L) + xVn−1,1 (Kλ , M ) ≤ 1
EQUALITY CONDITIONS OF THE BRUNN-MINKOWSKI THEOREM
3725
for all x ∈ [0, 1]. Since Vn (Kλ ) = 1, it follows from the inequality (5) that
f (x) = Vn ((1 − x)L + xM )
= Vn ((1 − x)L + xM ) Vn (Kλ )n−1
≤ Vn−1,1 (Kλ , (1 − x)L + xM )n ≤ 1,
for all x ∈ [0, 1]. Since f (0) = Vn (L) = 1, it follows that f (0) ≤ 0. On computing
f (0) as in (8), we have Vn−1,1 (L, M ) ≤ 1.
We have shown that if Vn−1,1 (Kλ , M ) ≤ 1, then Vn−1,1 (L, M ) ≤ 1. If M is a
singleton, then Vn−1,1 (Kλ , M ) = Vn−1,1 (L, M ) = 0. If M is not a singleton, then
Vn−1,1 (Kλ , M ) > 0, since Kλ has an interior. Let μ = Vn−1,1 (Kλ , M ). Then
Vn−1,1 (Kλ , μ1 M ) =
1
μ Vn−1,1 (Kλ , M )
= 1,
so that
1
μ Vn−1,1 (L, M )
= Vn−1,1 (L, μ1 M ) ≤ 1 =
1
μ Vn−1,1 (Kλ , M ).
In other words, Vn−1,1 (L, M ) ≤ Vn−1,1 (Kλ , M ) for all M .
But the argument above can be repeated, reversing the roles of Kλ and L. To
see this, recall the identity (11). It follows from (11) that for any value of λ, we
have equality in the Brunn-Minkowski inequality for convex combinations of Kλ
and L. Fixing a value of λ, and setting Sτ = (1 − τ )L + τ Kλ , apply the same
arguments as above using L and Kλ as the new “endpoints” to conclude that
Vn−1,1 (Kλ , M ) ≤ Vn−1,1 (Sτ , M ) for all M and all τ ∈ [0, 1]. When τ = 0 this
implies that Vn−1,1 (Kλ , M ) ≤ Vn−1,1 (L, M ) for all M . It now follows that
Vn−1,1 (Kλ , M ) = Vn−1,1 (L, M )
for all M and all λ ∈ [0, 1].
If n = 2, set λ = 0. Proposition 2.1 then implies that K and L are translates.
This case is the starting point for induction on the dimension n.
If n ≥ 3, then assume the theorem holds in lower dimensions. If u is a unit
vector, let M be the line segment ou so that
Vn−1,1 (Kλ , ou) = Vn−1,1 (L, ou).
It follows from (4) that
Vn−1 ((Kλ )u ) = Vn−1 (Lu )
for all λ ∈ [0, 1] so that
Vn−1 (Ku ) = Vn−1 ((1 − λ)Ku + λLu ) = Vn−1 (Lu ).
Therefore,
1
1
1
Vn−1 ((1 − λ)Ku + λLu ) n−1 = (1 − λ)Vn−1 (Ku ) n−1 + λVn−1 (Lu ) n−1 .
This is the equality case of the Brunn-Minkowski inequality in dimension n − 1.
Since K and L have interior, so do their projections (relative to (n − 1)-dimensional
subspaces). It now follows from the induction hypothesis that Ku and Lu are
homothetic. Since K and L have homothetic projections in every direction u, it
follows from Theorem 3.1 that K and L are homothetic.
3726
DANIEL A. KLAIN
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Department of Mathematical Sciences, University of Massachusetts Lowell,
Lowell, Massachusetts 01854
E-mail address: Daniel [email protected]